3 Answers
3

Your second equation, $P(\nu,T) = \frac{2 h {\nu}^3}{c^2}$ $\frac{1}{\exp\bigl(\frac{h \nu}{kT}\bigr) - 1}$ is what is commonly referred to as Planck's law for radiation, although a more standard symbol used is $B_\nu(T)$. This is the energy radiated per time, per area, per frequency interval, per steradian. It is a formula for the 'specific intensity' of a source, which intuitively is the energy flux along a ray of radiation in a given direction, and so you must normalize by the solid angle subtended by that ray.

To get the total energy per time per area radiated by a patch of a black body, integrate over solid angle and over frequency. Be careful performing the solid angle integral, however, because you must include the geometric factor $\cos \theta$ that accounts for the projected area of the patch ($\theta = 0$ corresponds to a ray emitted in the normal direction). Rays leaving one side of a patch can only be directed into the upper hemisphere of the solid angle sphere. So the solid angle integral looks like this:

The $2 \pi$ out in front is for the azimuthal angle. Here, $F_\nu$ is what is commonly referred to as the specific flux ('specific' because it's still per unit frequency interval). Then, either by reading up on the Riemann $\zeta$ function, or just using a computer to tell you the answer, you can perform the frequency integral and get

$$ F = \sigma \, T^4$$

Here $F$ is what we commonly think of as the flux (energy per area per time), and $\sigma$ is the Stefan-Boltzmann constant,

The original poster asked about Planck's law, which refers to energy radiated per time per area per frequency interval per solid angle. In that case the second formula is correct. I've included more details in my answer
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kleingordonMar 10 '12 at 4:39

Ah, good catch there. (Though it wouldn't surprise me much if what the OP means by Planck's law is not exactly the standard meaning.)
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David Z♦Mar 10 '12 at 5:00

That's a catch indeed! @kleingordon answer is definitely more useful although I must say that it is not clear from the question what normalization OP implies.
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SlaviksMar 10 '12 at 5:50